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Logarithm and Antilogarithm Test 21

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Logarithm and Antilogarithm Test 21
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  • Question 1
    1 / -0
    If the approximate value of $$ \log _{ 10 }{ (4.04) }$$is $$0. \ abcdef,$$ it is given that $$\log _{ 10 }{ (4) }=0.6021$$ &$$ \log _{ 10 }{ (e) }=0.4343,$$ then the value of abcd must be
    Solution

  • Question 2
    1 / -0
    If $$\log_{10}e=0.4343$$, then $$\log_{10}1016$$ is
    Solution
    $$\log_{10}1016\Rightarrow \dfrac{\log 1016}{\log 10}$$
    $$\Rightarrow \dfrac{3.006949}{1}$$
    $$\rightarrow$$ Option $$C$$ is correct
  • Question 3
    1 / -0
    The number of zeroes after decimal and before first significant digit in $$(50)^{-100}$$ is equal to : (take $$log_{10}$$ 5=0.699)
    Solution
    So, the correct answer is '169'.

  • Question 4
    1 / -0
    $$2^{74}-2^{73}-2^{72}$$ is same as
    Solution
    $$2^{74}-2^{73}-2^{72}=2^{72}(2^2-2-1)=2^{72}\times 1=2^{72}$$
    Hence,correct option is $$A.$$
  • Question 5
    1 / -0
    Find the value of $$\log_{10}{\left(0.\bar{9}\right)}$$
    Solution
    To find value of $$\log_{10}{0.\bar9}$$

    Let $$x=0.\bar{9}=0.999999...$$

    $$\Rightarrow 10x=9.99999....$$

    $$\Rightarrow 10x-x=9$$

    $$\Rightarrow 9x=9$$

    $$\Rightarrow x=\dfrac{9}{9}=1$$

    $$\therefore x=1$$

    Let $$y=\log_{10}{1}$$

    $$\Rightarrow 1={10}^{y}$$

    $$\Rightarrow {10}^{y}={10}^{0}$$                 (since $${10}^{0}=1$$)

    Since bases are same we can equate the powers

    $$\therefore y=0$$

    Hence, $$\log_{10}{\left(0.\bar{9}\right)}=0$$

  • Question 6
    1 / -0
    If $$\log 4=1.3868$$, then the approximate value of $$\log\, (4.01)$$
    Solution
    Let $$y=f(x)=\log x$$

    Let $$x=4$$

    $$X+\triangle x=4.01$$

    $$\triangle x=0.01$$

    For $$x=4$$

    $$Y=\log4=1.3868$$

    $$y=\log x$$

    $$\dfrac{dy}{dx}=\dfrac{1}{x}=\dfrac{1}{4}$$

    $$\triangle y=dy$$

    $$=\dfrac{dy}{dx}.dx$$

    $$=\dfrac{1}{4}\times 0.01$$

    $$\triangle y=0.0025$$

    $$\log(4.01)=y+\triangle y$$$$=1.3893$$
  • Question 7
    1 / -0
    If $$x=500,y=100$$ and $$z=5050$$, then the value of $$(\log _{ xyz }{ { x }^{ z } } )(1+\log _{ x }{ yz } )$$ is equal to.
    Solution
    Given,

    $$\left(\log _{xyz}\left(x^z\right)\right)\left(1+\log _x\left(yz\right)\right)$$

    $$\left(\log _{xyz}\left(x^z\right)\right)\left(1+\log _x\left(yz\right)\right)$$

    $$=z\log _{xzy}\left(x\right)\left(\log _x\left(zy\right)+1\right)$$

    from given w have,

    $$=5050\log _{(500 \times 5050 \times 100)}\left(500\right)\left(\log _{500}\left((5050 \times 100)\right)+1\right)$$

    $$=\dfrac{5050\log_e \left(505000\right)}{\log_e \left(252500000\right)}+5050\log _{252500000}\left(500\right)$$

    $$=5050$$

  • Question 8
    1 / -0
    The greatest value of $$(4\log_{10}{x}-\log_{x}{(0.0001)})$$ for $$0 < x < 1$$ is
    Solution
    $$(4log_{10}x-log_x(0.0001)$$
    $$(4log_{10}x-log_{x}\dfrac{1}{10^4}$$
    $$(4log_{10}x-(log_x1-log_x{10^4})$$
    $$(4log_{10}x-0+log_x{10^4})$$
    $$(4log_{10}x+log_x{10^4})$$
    $$4(log_{10}x+log_x10)$$
    $$4(log_{10}x+\dfrac{1}{log_{10}x})$$
    $$AM\geq GM$$
    $$\dfrac{x+y}{2} \geq \sqrt{xy}$$
    $$log_{10}x+\dfrac{1}{log_{10}x} \geq 2$$
    $$4(log_{10}x+\dfrac{1}{log_{10}x}) \geq 8$$


  • Question 9
    1 / -0
    The value of $$ 3 ^{log_4 5} -5 ^{log_4 3}$$
    Solution

  • Question 10
    1 / -0
    The number of $$\log_2 7 $$ is 
    Solution

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